A2-4a + 9b2 + 6B + 5 = 0 for ab-b-a

A2-4a + 9b2 + 6B + 5 = 0 for ab-b-a

a2-4a+9b2+6b+5=0
(a^2-4a+4)+(9b^2+6b+1)=0
(a-2)^2+(3b+1)^2=0
The sum of two nonnegative numbers is 0, and each nonnegative number is 0
So a = 2, B = - 1 / 3
So (AB) / (B-A) = - 2 / 3 / (- 7 / 3) = 2 / 7

Given a2-4a + 9b2 + 6B + 5 = 0, find the value of a + B
I looked at the answer. It's 3 / 5

The original formula can be decomposed into (A-2) ^ 2 + (3b + 1) ^ 2 = 0, the sum of two bisections is zero, so each term is 0
That is, a = 2, B = - 1 / 3, so a + B = 5 / 3

It is known that aa-4a + 9bb + 6B + 5 = 0, and the value of 1 / A-1 / B is obtained

2A + 3B = 1, 4a-6b = how much
RT
Don't tell me it's equal to - 1

1

If a and B are real numbers, and A2 + 4b2-2a + 4B + 2 = 0, the value of 1 / 4a2 + B is required
A2 and B2 both mean square

Simplified radical (a2-4a + 4) - radical (A2 + 4A + 4)

a2- 4a+4=(a-2)^2
a2+4a+4=(a+2)^2
therefore
Under radical (a2-4a + 4) - under radical (A2 + 4A + 4)
=a-2-(a+2)
=4

[a2-4ab+4b2]-[2a-4b]+1

Factorization A2 + 2A + 4B2 + 4B + 4ab-3

If A2 + a = 0, then the value of 2A2 + 2A + 2007 is______ ．

∵ A2 + a = 0, ∵ 2A2 + 2A = 0, substituting 2A2 + 2A = 0, then 2A2 + 2A + 2007 = 2007

Given A2 + 2A -- 1 = 0, find the value of A4 + 2A2 + 16A -- 1
2 after a is quadratic
4 after a is the fourth power

6
It is known that A2 + 2A = 1
And A2 = 1-2a
a4+2a2+16a-1
=(1-2a)2+2(1-2a)+16a-1
=4a2+8a+2
=4(a2+2a)+2
=4*1+2
=6